A ball is thrown straight up from the top of a 24 foot tall building with an initial velocity of
40 feet per second. The height of the ball as a function of time can be modeled by the function
h = -16t2 + 40t + 24
Part A
What is the height of the ball after 1 second? h=
Part B
How long will it take for the ball to hit the ground (set h=0 and factor completely)
seconds

Step-by-step explanation:

Part A

The height of the ball as a function of time can be modeled by the function

h = -16t² + 40t + 24

When t = 1 second,

h = - 16 × 1² + 40 × 1 + 24

h = - 16 + 40 + 24

h = 48 feet

Part B

The equation would be

h = -16t² + 40t + 24 = 0

Dividing both sides of the equation by 4, it becomes

- 4t² + 10t + 6 = 0

We would find two numbers such that their sum or difference is 10t and their product is - 24t².

The two numbers are 12t and - 2t. Therefore,

- 4t² + 12t - 2t + 6 = 0

- 4t(t - 3) - 2(t - 3) = 0

t - 3 = 0 or - 4t - 2 = 0

t = 3 or t = 2/- 4 = - 1/2

Since the time cannot be negative, then t = 3 seconds

It will take 3 seconds for the ball to hit the ground.

I think the answer is B